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Homework Help: Hausdorff Space

  1. Jan 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that X is Hausdorff if and only if the diagonal [tex]\Delta = \{x \times x | x \in X \}[/tex] is closed in [tex]X \times X[/tex].

    2. Relevant equations

    Definition of Hausdorff Space (T2) : A topological space in which distinct points have disjoint neighborhoods.

    3. The attempt at a solution
     
  2. jcsd
  3. Jan 24, 2009 #2
    With so little work on your part shown to go by it's difficult to know where you're stuck.

    If you haven't already, express the closedness of [tex]\Delta[/tex] in [tex]X\times X[/tex] (which I assume has the product topology) in terms of the openness of its complement.

    Now have a go at proving each direction (neither is more difficult than the other) if you haven't already, and please show us your efforts.
     
  4. Jan 24, 2009 #3
    ->
    If X is Hausdorff, the diagonal [tex] \Delta [/tex] is closed in [tex]X\times X[/tex].

    Assume X is Hausdorff. Now, we have two distinct points x, y and disjoint open sets U, V containing x, y, respectively. The basis element [tex]U \times V[/tex] containing [tex](x,y) \in X \times X[/tex] should not intersect [tex] \Delta [/tex] by the assumption given to the Hausdorff property.
    For every [tex] (x,y) \notin \Delta [/tex], we have a basis element in [tex] X \times X [/tex] containing (x,y), which does not intersect [tex] \Delta [/tex].
    Thus, [tex]X \times X \setminus \Delta is open[/tex] and we conclude [tex] \Delta [/tex] is a closed set in [tex]X \times X[/tex] .

    <-
    If the diagonal [tex] \Delta [/tex] is closed in [tex]X\times X[/tex], X is Haudorff.

    Supppose [tex] \Delta [/tex] is closed in [tex]X\times X[/tex]. Then, [tex]X \times X \setminus \Delta [/tex] is open. Let [tex](x,y) \in X \times X[/tex] and [tex] x \neq y [/tex]. For [tex](x,y) \notin \Delta[/tex], we have a basis element [tex] U \times V [/tex] in [tex]X \times X [/tex] containing (x, y).
    We remain to show U and V are disjoint. Suppose on the contrary that U and V are not disjoint. Then, there is an element [tex] (z,z) \in X times X [/tex] which belongs to both U and V. Contradicting the fact that x and y are distinct.
    Thus, X is Hausdorff.
     
  5. Jan 24, 2009 #4
    That's quite a leap from your previous post.

    Now how 'bout you tackle your https://www.physicsforums.com/showthread.php?t=287038" without copying down the solution from an external source. It's the only way you'll learn topology.
     
    Last edited by a moderator: Apr 24, 2017
  6. Jan 24, 2009 #5
    Can you please give me a link of an external source you mentioned?
    It's my self study of topology (I am not even majoring in math) and I don't need to copy the external source to show you something to impress you. Rather, I just did for my self study purpose and asked an advice if someone finds an error in my attempt to the solution.
     
  7. Jan 24, 2009 #6
    Well, in that case, your work is quite error-free indeed! I would only suggest rephrasing picking an element [tex](x,y)\not\in \Delta[/tex] as picking an element [tex](x,y)\in (X\times X)\backslash \Delta[/tex]. Now, on to your other problem!
     
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